The emission spectra is observed by the consequence of transition of electron from higher energy state to ground state of $H e^{+}$ ion. Six different photons are observed during the emission spectra, then what will be the minimum wavelength during the transition?

\frac{4}{27 R_{H}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{4}{15 R_{H}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

λ = \frac{15}{16 R_{H}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{16}{15 R_{H}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{4}{15 R_{H}}$
According to given data:
$6 = (n_{2} - n_{1}) \frac{(n_{2} - n_{1} + 1)}{2}; n_{2} = 4; n_{1} = 1$
$∴ \frac{1}{λ} = R_{H} \times 2^{2} [1 - \frac{1}{16}]$
$\Rightarrow R \times \frac{15}{4}; λ = \frac{4}{15 R_{H}}$

Suggest Corrections

Similar questions

During the emission spectrum of electron in $H e^{+}$ ion, the electron on transition from same higher energy state drops to the ground state and first excited state emits two photon of $λ_{1}$ =304 $A^{\circ}$ and $λ_{2}$ =1085 $A^{\circ}$ respectively.The total number of photons emitted during the transition of electrons from highest excited to all lower energy level(Possible)